The digits indicated by ^* in 3422213^{**} so | vedclass

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(A) Let the number be $N = 3422213xy$,where $x$ and $y$ are the missing digits.For $N$ to be divisible by $99$,it must be divisible by both $9$ and $1

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$1, 9$

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(A) Let the number be $N = 3422213xy$,where $x$ and $y$ are the missing digits.For $N$ to be divisible by $99$,it must be divisible by both $9$ and $11$.For divisibility by $9$,the sum of the digits must be a multiple of $9$:$3+4+2+2+2+1+3+x+y = 17+x+y$. For this to be divisible by $9$,$x+y$ must be $1$ or $10$ or $19$. Since $x, y$ are digits,$x+y=10$ (as $17+10=27$,which is divisible by $9$).For divisibility by $11$,the difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$:Sum of digits at odd places (from right): $y+1+2+4 = y+7$.Sum of digits at even places (from right): $x+3+2+2+3 = x+10$.Difference: $(y+7) - (x+10) = y-x-3$.For divisibility by $11$,$y-x-3 = 0$ or $y-x-3 = -11$.Case $1$: $y-x = 3$. We have $x+y=10$ and $y-x=3$. Adding these,$2y=13$ (no integer solution).Case $2$: $y-x-3 = -11 \implies y-x = -8$. We have $x+y=10$ and $y-x=-8$. Adding these,$2y=2 \implies y=1$. Then $x=9$.Thus,the digits are $9$ and $1$.

The digits indicated by $^*$ in $3422213^{**}$ so that this number is divisible by $99$ are

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